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# NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions

In this exercise 3.4 of the trigonometric function, we will study the solutions of trigonometric equations. The  equation containing trigonometric functions are known as trigonometric equations,like sinx = siny, cosx +1 =0 etc. The solutions of these trigonometric equations are of two kinds.

1-Principal Solution. The values of an unknown angle between 0° and 360°that satisfy the given equation are known as principal solutions of the given equation.

2-General Solutions. Set of infinite solutions that satisfy the given equation is known as the general solution, this solution is lead due to the periodicity of the trigonometric functions. The general solution depends upon period of the trigonometric function, as an example if principal solution of the equation sinx =1 ⇒ sinx= sinπ/2,so principal solution is x =π/2 then, this equation will satisfy for x= (2π ±π/2), (4π± π/2)…..etc, so in compact form, we can write these solutions x =nπ +(–1)n×π/2, it is all about the general solution. The periodicity and general solutions of trigonometric functions are given below.

### CBSE Class 11-Question paper of maths 2015

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## NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions

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Q1. Find the  principal and general solution of the equation tanx = √3.

Hint: All possible values of x lying between 0 and 2π which satisfy the equation are called principal solutions.

Therefore, substituting √3 by tanπ/3 and tan 4π/3 we get

For calculating general solution if we have tanx = tany then the value of x = nπ + y is known as general solution of the equation tanx = tany,so

Taking the smaller value of the solution π/3

4π/3 was not taken here because it is one of the infinite solutions ( for n=1, x = π +π/3 =4π/3)

(n∈ z)

will be the general solution of the equation.

Q2. Find the principal and general solution of the equation secx = 2.

secx =2

Since the value of secx is positive(i.e 2),x will lie on 1 st and 4 th quadrant.(value of secx is positive in 1st and 4 th quadrant).

4 th quadrant means 2π −x (x =π/3)

Therefore principal solutions of the equation will be

If secx = secy then  general solution of the equation is given by the same expression as for the cosx =cosy, see the hint.

Hint: If cosx = cosy⇒1/secx = 1 /secy⇒secx=secy,therefore secx =secy implies that cosx = cosy.

(n ∈ z)

Therefore the general solution of the equation will be 2nπ ± π/3.

Q3. Find the principal and general solutions of the equation cotx = –√3.

cotx = –√3.

√3 = cotπ/6

Value of cotx is negative ,so  x  will lie on 2nd quadrant (i.e π– π/6) and fourth quadrant(i.e 2π– π/6)(value of cotx is negative in 2 nd and 4 th quadrant).

Therefore the principal solutions of the given equation will be,

Therefore  the general solution of the given equation is

Q4.Find the general solution of cosecx =–2.

Value of cosec is negative in 3rd(π+π/6) and 4 th quadrant (2π–π/6).

Therefore the principal solutions of the given equation will be,

The general solution of the given equation will be

As we know cosecx = 1/sinx

Q5.Find the general solution of the equation cos4x = cos2x.

cos4x = cos2x

cos4x – cos2x = 0

⇒2sin3x.sinx= 0

⇒sin3x=0 and sinx =0

sin3x=0

sin3x = sin0

since θ = 0

3x = nπ

sinx =0 =sin0

x = nπ

Therefore general solution of both the equation will be

Q6. Find the general solution of the equation cos3x + cosx – cos2x = 0.

cos3x + cosx – cos2x = 0

(cos3x + cosx )– cos2x = 0

⇒2cos2x.cosx – cos2x =0

⇒cos2x(2cosx – 1) = 0

Since general solution of the equation cosx = cosy is x = 2nπ ± y,so

cos2x = 0

2cosx – 1 =0

2cosx = 1

Therefore the general solution of the given equation will be

Q7. Find the general solution of the equation sin2x + cosx = 0.

sin2x + cosx = 0

2sinx.cosx + cosx = 0

cosx(2sinx + 1) = 0

cosx = 0, 2sinx + 1 = 0

cosx = cos0, sinx = –1/2⇒sinx = -sinπ/6 = sin(π+ π/6)=sin7π/6

Therefore the general solution of the given equation will be ,

Q8. Find the general solution of the  equation sec²2x = 1 – tan2x.

sec²2x = 1 – tan2x

1 + tan²2x  = 1 – tan2x

tan²2x + tan2x = 0

tan2x(tan2x +1) = 0

tan2x = 0

tan2x = tan0( general solution of the equation (tanx = tany) is x = nπ + y)

2x = nπ + 0

tan2x +1 = 0

tan2x = –1

Value of tan is negative so 2x will lye in 2nd(π –π/4) and 4 th quadrant (2π – π/4)[tanθ is negative in 2nd and 4 th quadrant].

Taking the smaller value of 2x

Therefore the general solution of the given equation is

(n εz)

Q9. Find the general solution of the equation sinx + sin3x + sin5x = 0.

sinx + sin3x + sin5x = 0

(sinx + sin5x )+ sin3x = 0

⇒2sin3x.cos(–4x) + sin3x=0

⇒2sin3x.cos4x +sin3x=0

⇒sin3x(2cos4x +1) =0

sin3x = 0 and 2cos4x +1 =0⇒cos4x = –1/2=–cosπ/3

cos4x = –cosπ/3

Value of cos4x  is negative in 2nd (π -π/3) and 3rd quadrant(π + π/3),taking smaller value in getting general solution.

Therefore the general solution of the given equation is

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### 1 thought on “NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions”

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